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labwork [276]
2 years ago
9

6. During some actual expansion and compression processes in piston–cylinder devices, the gases have been

Engineering
1 answer:
Katyanochek1 [597]2 years ago
5 0

During some actual expansion and compression processes in piston-cylinder devices, the gases have been are the P1= P2.

<h3>What is the pressure?</h3>

Pressure is something that has the pressure that is physical and that causes the pressure is piston-cylinder devices.

During a few real enlargements and compression procedures in piston-cylinder devices, the gases were located to meet the connection PV n = C, wherein n and C are constants.

Read more about the pressure :

brainly.com/question/25736513

#SPJ1

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. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and si
Over [174]

Answer:

y  ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

v =1/n R^{2/3} s^{1/2}

where R is hydraulic radius

S = bed slope

Q = Av =A 1/n R^{2/3} s^{1/2}

A = 1/2 \times (B+B+4y) \times y =(B+2y) y

R =\frac{A}{P}

P is perimeter =  (B+2\sqrt{5} y)

R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}

Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}

solving for y100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}

solving for y value by using iteration method ,we get

y  ≈ 2.5

5 0
3 years ago
Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale
artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

  • HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ \longrightarrow 2Cl₂ + 2H₂O

Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \  of \ limiting \ reactant \ supplied \ in \ the \, feed}}

Which gives;

80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

  • <u>HCl: 0.4 kmol/h</u>
  • <u>Cl₂: 1.6 kmol/h</u>
  • <u>H₂O: 1.6 kmol/h</u>
  • <u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

brainly.com/question/14668990

brainly.com/question/82989

7 0
3 years ago
A flow rate sensing device used on a liquid transport pipeline functions as follows. The device provides a 5-bit output where al
marysya [2.9K]

Answer:

Explanation:

The step by step analysis is as shown in the attached files.

8 0
3 years ago
An angle is observed repeatedly using the same equipment and procedures producing the data below:35 ∘ 40'00",35 ∘ 40'10",35 ∘ 40
Helen [10]

Answer: (a) +/- 7.5° (b) +/- 3.75°

Explanation:

See attachment

6 0
3 years ago
Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.
Sholpan [36]

Answer:

Frequency = 3.5294\times 10^{14}s^{-1}

Wavenumber = 1.1765\times 10^6m^{-1}

Energy = 2.3365\times 10^{-19}J

Energy = 1.4579 eV

Energy = 2.3365\times 10^{-12}erg

Explanation:

As we are given the wavelength = 850 nm

conversion used : (1nm=10^{-9}m)

So, wavelength is  850\times 10^{-9}m

The relation between frequency and wavelength is shown below as:

Frequency=\frac{c}{Wavelength}

Where, c is the speed of light having value = 3\times 10^8m/s

So, Frequency is:

Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}

Frequency=3.5294\times 10^{14}s^{-1}

Wavenumber is the reciprocal of wavelength.  

So,  

Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}

Wavenumber=1.1765\times 10^6m^{-1}

Also,  

Energy=h\times frequency

where, h is Plank's constant having value as 6.62\times 10^{-34}J.s

So,  

Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})

Energy=2.3365\times 10^{-19}J

Also,  

1J=6.24\times 10^{18}eV

So,  

Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)

Energy=1.4579eV

Also,  

1J=10^7erg

So,  

Energy=(2.3365\times 10^{-19})\times 10^7erg

Energy=2.3365\times 10^{-12}erg

5 0
3 years ago
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